00:01
Hello students, in order to show xi of x as a solution, let us consider for free particle this schrodinger equation is given by minus h cross square divided by 2m d square xi of x by dx square plus v of x into xi of x which is equal to e into xi of x.
00:31
Here for free particle we can take v of x is equal to 0, then the above equation become h cross square by 2m d square by dx square of xi of x minus e xi of x equal to 0.
00:53
Here multiply and divided by 2m h cross square so that we get d square xi of x divided by dx square plus 2m e divided by h cross square xi of x.
01:16
Here our main term is to make before this term there should be no term is present.
01:26
Hence, whatever the term present behind this d square by dx square we need to divide by that.
01:32
So when we divide that this term here comes as 2m by h cross square.
01:39
So now take this 2m e by h cross square as k that is constant.
01:47
Now by taking constant term we can write k square xi of x is equal to 0.
01:59
Hence k will become root of 2m e by h cross square.
02:07
Here d square plus k square is equated to 0 that is this term d square plus k square.
02:17
From this d will be equal to plus or minus i into k that is the solution of differential equation which is having d is equal to plus or minus ik so that we can write xi of x is equal to a cos kx plus b sin kx.
02:43
If the function is that we can get the solution as plus or minus ik.
02:49
Hence the proof for the first question...